Draw the graph of `f(x)=|(x^2-2)/(x^2-1)|` .

Explanation

Let us first draw the graph of the function `g(x)=(x^(2)-2)/(x^(2)-1)`.
The domain of the function is `R-{-1,1}`.
1. y-intercept
`f(0)=2`
So the graph cuts the y-axis at (0,2).
2. x-intercept (zeros)
Put `y=0` or `x^(2)-2=0 :. x=+-sqrt(2)`
Hence the graph intersects the x-axis at `(+-sqrt(2),0)`.
3. Asymptotes
Vertical asymptotoes
Graph has the vertical asymptote as `x^(2)-1=0` or `x=+-1`.
Horizontal asymptotes
Since the degree of the numerator and that of the denominator are same, the horizontal asymptote is y=ratios of leading coefficients or y=1.
Clearly, the graph has no oblique asymptote.
4. Extremum
`f'(x)=((2x)(x^(2)-1)-(x^(2)-2)2x)/((x^(2)-1)^(2))=(2x)/((x^(2)-1)^(2))`
`f'(x) gt 0 ` for `xgt0, x!=1` where the function increases.
`f'(x)lt0` for `xlt0, x!=-1` where the function decreases.
`f'(x)=0 rArrx=0`, where is the point of minima.
Thus, important points and lines are as follows.

`underset(xrarr-oo)lim(x^(2)-2)/(x^(2)-1)=1` and `underset(xrarr-1^(-1))lim(x^(2)-2)/(x^(2)-1)=-oo`
So f(x) decreases from 1 to `-oo` when x increases from `-oo` to -1 crossing the x-axis at `(-sqrt(2),0)`.
`underset(xrarr-1^(+))lim(x^(2)-2)/(x^(2)-1)=oo` and `f(0)=2`
So f(x) decreases from `oo` to 2 when x increases from -1 to 0.
`underset(xrarr1^(-))lim(x^(2)-2)/(x^(2)-1)=oo`
So f(x) increases from 2 to `oo` when x increases from 0 to 1.
`underset(xrarr1^(+))lim(x^(2)-1)/(x^(2)-1)=-oo` and `underset(xrarroo)lim (x^(2)-2)/(x^(2)-1)=1`
So f(x) increases form `-oo` to 1 when x increases from 1 to `oo` crossing the x-axis at `(sqrt(2),0)`.
Hence the graph of the function can be drawn as shown in the following figure.

Also the function is even, hence the graph is symmetrical about the y-axis.
To draw the graph of `y=f(x)=|(x^(2)-2)/(x^(2)-1)|`, flip the portion of the graph of `y=(x^(2)-2)/(x^(2)-1)` which lies below the x-axis over the x-axis.